What is the answer to p=((2^{n+4}+2(2^n)))/((3(2^{n+3)))} ?

The answer to p=((2^{n+4}+2(2^n)))/((3(2^{n+3)))} is p= 3/4

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p=((2^{n+4}+2(2^n)))/((3(2^{n+3)))}

p = \frac{( 2 ^{n + 4} + 2 ( 2 ^{n} ) )}{( 3 ( 2 ^{n + 3} ) )}

p = \frac{3}{4}

Decimal Notation

p = 0.75

Solution steps

p = \frac{( 2 ^{n + 4} + 2 ( 2 ^{n} ) )}{( 3 ( 2 ^{n + 3} ) )}

Simplify

\frac{2 ^{n + 4} + 2 ( 2 ^{n} )}{3 ( 2 ^{n + 3} )} : \frac{3}{4}

p = \frac{3}{4}

z = (35 + 18 i) + (- 10 - 3 i)

x, x - \frac{y}{5} - 10 = 0

t, a (t) = a (1 + r)

x, 5 (1_{2} t) + 2 (3 - x) = 3 (x + 4) + 14

y, tan (x) - 2 x^{3} \cdot y = c

What is the answer to p=((2^{n+4}+2(2^n)))/((3(2^{n+3)))} ?
The answer to p=((2^{n+4}+2(2^n)))/((3(2^{n+3)))} is p= 3/4